Pre-Algebra Examples

$\frac{{(x - 5)}^{2}}{25} + \frac{{(y - 2)}^{2}}{16} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(x - 5)}^{2}}{25} + \frac{{(y - 2)}^{2}}{16} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = 5$

$b = 4$

$k = 2$

$h = 5$

Step 4

The center of an ellipse follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(5, 2)$

Step 5

Find $c$ , the distance from the center to a focus.

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Step 5.1

Find the distance from the center to a focus of the ellipse by using the following formula.

$\sqrt{a^{2} - b^{2}}$

Step 5.2

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(5)}^{2} - {(4)}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

Raise $5$ to the power of $2$ .

$\sqrt{25 - {(4)}^{2}}$

Step 5.3.2

Raise $4$ to the power of $2$ .

$\sqrt{25 - 1 \cdot 16}$

Step 5.3.3

Multiply $- 1$ by $16$ .

$\sqrt{25 - 16}$

Step 5.3.4

Subtract $16$ from $25$ .

$\sqrt{9}$

Step 5.3.5

Rewrite $9$ as $3^{2}$ .

$\sqrt{3^{2}}$

Step 5.3.6

Pull terms out from under the radical, assuming positive real numbers.

$3$

Step 6

Find the vertices.

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Step 6.1

The first vertex of an ellipse can be found by adding $a$ to $h$ .

$(h + a, k)$

Step 6.2

Substitute the known values of $h$ , $a$ , and $k$ into the formula.

$(5 + 5, 2)$

Step 6.3

Simplify.

$(10, 2)$

Step 6.4

The second vertex of an ellipse can be found by subtracting $a$ from $h$ .

$(h - a, k)$

Step 6.5

Substitute the known values of $h$ , $a$ , and $k$ into the formula.

$(5 - (5), 2)$

Step 6.6

Simplify.

$(0, 2)$

Step 6.7

Ellipses have two vertices.

${Vertex}_{1}$ : $(10, 2)$

${Vertex}_{2}$ : $(0, 2)$

${Vertex}_{1}$ : $(10, 2)$

${Vertex}_{2}$ : $(0, 2)$

Step 7

Find the foci.

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Step 7.1

The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

Step 7.2

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(5 + 3, 2)$

Step 7.3

Simplify.

$(8, 2)$

Step 7.4

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Step 7.5

Substitute the known values of $h$ , $c$ , and $k$ into the formula.

$(5 - (3), 2)$

Step 7.6

Simplify.

$(2, 2)$

Step 7.7

Ellipses have two foci.

${Focus}_{1}$ : $(8, 2)$

${Focus}_{2}$ : $(2, 2)$

${Focus}_{1}$ : $(8, 2)$

${Focus}_{2}$ : $(2, 2)$

Step 8

Find the eccentricity.

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Step 8.1

Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} - b^{2}}}{a}$

Step 8.2

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(5)}^{2} - {(4)}^{2}}}{5}$

Step 8.3

Simplify the numerator.

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Step 8.3.1

Raise $5$ to the power of $2$ .

$\frac{\sqrt{25 - 4^{2}}}{5}$

Step 8.3.2

Raise $4$ to the power of $2$ .

$\frac{\sqrt{25 - 1 \cdot 16}}{5}$

Step 8.3.3

Multiply $- 1$ by $16$ .

$\frac{\sqrt{25 - 16}}{5}$

Step 8.3.4

Subtract $16$ from $25$ .

$\frac{\sqrt{9}}{5}$

Step 8.3.5

Rewrite $9$ as $3^{2}$ .

$\frac{\sqrt{3^{2}}}{5}$

Step 8.3.6

Pull terms out from under the radical, assuming positive real numbers.

$\frac{3}{5}$

Step 9

These values represent the important values for graphing and analyzing an ellipse.

Center: $(5, 2)$

${Vertex}_{1}$ : $(10, 2)$

${Vertex}_{2}$ : $(0, 2)$

${Focus}_{1}$ : $(8, 2)$

${Focus}_{2}$ : $(2, 2)$

Eccentricity: $\frac{3}{5}$

Step 10